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For the following variational problem I have been told to show the Euler-Lagrange equation is identically zero.

$$L[u]:= \int_a^b \sin(u)u_x\,\mathrm dx $$

I have found it to be

$u_x\cos(u)-\sin(u)u_{xx}=0.$

Is this correct? And if so, does this always equal $0$?

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  • $\begingroup$ I think your second term is wrong, it should be something like $\frac{d}{dx} \sin(u)$, which will end up canceling the first term. $\endgroup$ – asperanz Oct 21 '12 at 20:01
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No: write $F(t_1,t_2,t_3):=t_3\sin t_2$. Then $L(u)=\int_a^bf(t,u(t),u'(t))dt.$ Euler-Lagrange equation is $\partial_{2}F(t,u,u')=\frac d{dt}\partial_3F(t,u,u')$, hence $$u'(t)\cos u(t)-\frac d{dx}\sin(u(t))=0, $$ which is always satisfied.

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Your equation of motion is not correct, you were already given the right one. But indeed your equation is trivially satisfied. That's because the Lagrangian itself is a total derivative, then the action

$$L[q]=-\int_a^b\frac{d}{dt}\cos(q(t))\mathrm{d}t=cte.$$

does not depend on the path $q(t)$ but on the endpoints, which you held fixed. That is, as a variational problem, it is trivial as any Lagrangian having the form $\mathcal{L}[q,t]=\dot{f}(q(t))$. Actually, Lagrangians are not uniquelly, but defined up to those kind of terms.

In higher dimensions the Lagrangians that mimic that in your question are

$$\mathcal{L}[\phi]=\mathrm{d}{\phi}$$

for $\phi$ a (dim$M-1$)-form, so that, by Stokes theorem:

$$ S=\int_M\mathcal{L} =\int_M\mathrm{d}\phi = \int_{\partial M} \phi $$

If $M$ is boundaryless or $\phi$ is compactly supported, this term has trivial equations of motion $-$ no dynamics.

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